What is the extraneous solution to these equations? $\dfrac{x^2 + 1}{x + 2} = \dfrac{11x - 23}{x + 2}$
Answer: Multiply both sides by $x + 2$ $ \dfrac{x^2 + 1}{x + 2} (x + 2) = \dfrac{11x - 23}{x + 2} (x + 2)$ $ x^2 + 1 = 11x - 23$ Subtract $11x - 23$ from both sides: $ x^2 + 1 - (11x - 23) = 11x - 23 - (11x - 23)$ $ x^2 + 1 - 11x + 23 = 0$ $ x^2 + 24 - 11x = 0$ Factor the expression: $ (x - 8)(x - 3) = 0$ Therefore $x = 8$ or $x = 3$ The original expression is defined at $x = 8$ and $x = 3$, so there are no extraneous solutions.